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2 years ago

Section 1: Experimental Overview

Physics

1 answer:

Luden [163]2 years ago

7 0

Answer:

1-state what the lab is about, that is, what scientific concept (theory, principle, procedure, etc.) you are supposed to be learning about by doing the lab. You should do this briefly, in a sentence or two. If you are having trouble writing the opening sentence of the report, you can try something like: "This laboratory experiment focuses on X…"; "This lab is designed to help students learn about, observe, or investigate, X…." Or begin with a definition of the scientific concept: "X is a theory that…."

2-give the necessary background for the scientific concept by telling what you know about it (the main references you can use are the lab manual, the textbook, lecture notes, and other sources recommended by the lab manual or lab instructor; in more advanced labs you may also be expected to cite the findings of previous scientific studies related to the lab). In relatively simple labs you can do this in a paragraph following the initial statement of the learning context. But in more complex labs, the background may require more paragraphs.

Explanation:

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Consider a uniform solid sphere of radius R and mass M rolling without slipping. Which form of its kinetic energy is larger, tra

castortr0y [4]

Answer:

A. Its translational kinetic energy is larger than its rotational kinetic energy.

Explanation:

Given that

Radius = R

Mass = M

We know that mass moment of inertia for the solid sphere

$I=\dfrac{2}{5}MR^2$

Lets take angular speed =ω

Linear speed =V

Condition for pure rolling , V= ω R

Rotation energy ,RE

$RE=\dfrac{1}{2}I\omega^2$

$RE=\dfrac{1}{2}\times \dfrac{2}{5}MR^2\times \omega^2$

$RE=\dfrac{1}{5}\times MR^2\times \omega^2$

$RE=\dfrac{1}{5}\times MV^2$

RE= 0.2 MV²

The transnational kinetic energy TE

$TE=\dfrac{1}{2}MV^2$

TE= 0.5 MV²

From above we can say that transnational energy is more than rotational energy.

Therefore the answer is A.

3 0

3 years ago

In Speed Study Number 1, we looked at two cars traveling the same distance at different speeds on city streets. Car "A" traveled

Ganezh [65]

Answer:

230.4 s

Explanation:

The speed of car A is

$v_A = 35 mi/h$

and the distance travelled is

$d = 10 mi$

so the time taken for car A is

$t_A = \frac{d}{v_A}=\frac{10 mi}{35 mi/h}=0.286 h$

The speed of car B is

$v_B = 45 mi/h$

and the distance travelled is

$d = 10 mi$

so the time taken for car B is

$t_B = \frac{d}{v_B}=\frac{10 mi}{45 mi/h}=0.222 h$

So the difference in time is

$\Delta t = t_A - t_B = 0.286 h -0.222 h=0.064 h$

Which corresponds to

$\Delta t = 0.064 h \cdot 3600 s/h = 230.4 s$

so car B arrived 230.4 s before car A.

5 0

3 years ago

When hydrogen is fused into helium, energy is released from Choose one: A. the increase in pressure. B. the decrease in the grav

Tatiana [17]

The nuclear fusion of hydrogen atoms releases a huge amount of energy. So the correct choice is C. Conversion of mass to energy.

What is nuclear fusion?

When two small nuclei join to form a new nucleus, then this process is termed nuclear fusion. A huge amount of energy is released when there occurs nuclear fusion between the two nuclei. And a new element is formed.

It has been observed that the amount of energy released in nuclear fusion is equal to the mass difference between the mass of the formed nucleus and the total mass of old nuclei. Hence in the nuclear fusion of hydrogen nuclei to form a helium nucleus, the energy is released due to the conversion of mass into energy.

The pressure is increased to make the hydrogen atoms fuse but this change in pressure does not contribute to the energy released in the fusion of hydrogen.

The magnitude of the gravitational field is too low and it does not contribute to the energy released in the fusion of hydrogen.

The gravitational collapse does not occur between the two hydrogen atoms. This phenomenon occurs in celestial bodies so this also does not contribute to the energy released in the fusion of hydrogen.

Learn more about nuclear fusion here:

brainly.com/question/10165218

#SPJ4

8 0

1 year ago

"In a Young’s double-slit experiment, the separation between slits is d and the screen is a distance D from the slits. D is much

san4es73 [151]

Answer:

The number of bright fringes per unit width on the screen is, $x=\dfrac{\lambda D}{d}$

Explanation:

If d is the separation between slits, D is the distance between the slit and the screen and $\lambda$ is the wavelength of the light. Let x is the number of bright fringes per unit width on the screen is given by :